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Solve-sin-2x-cos-2x-0-for-pi-lt-x-lt-pi-




Question Number 15975 by myintkhaing last updated on 16/Jun/17
Solve sin 2x + cos 2x =0,  for −π<x<π.
$${Solve}\:{sin}\:\mathrm{2}{x}\:+\:{cos}\:\mathrm{2}{x}\:=\mathrm{0}, \\ $$$${for}\:−\pi<{x}<\pi. \\ $$
Commented by tawa tawa last updated on 16/Jun/17
sin2x + cos2x = 0  Note: sin2x = 2sinxcosx  and  cos2x = cos^2 x − sin^2 x  2sinxcosx + cos^2 x − sin^2 x = 0  cos^2 x = 1 − sin^2 x  2sinxcosx + 1 − sin^2 x − sin^2 x = 0  2sinxcosx + 1 − 2sin^2 x = 0  ∴ 2sin^2 x − 2sinxcosx − 1 = 0  ∴ 2sin^2 x − 2sinx(√(cos^2 x)) − 1 = 0  ∴ 2sin^2 x − 2sinx(√(1 − sin^2 x)) − 1 = 0  sinx = p  ∴  2p^2  − 2p(√(1 − p^2 )) − 1 = 0  ∴  2p^2  − 1 =  2p(√(1 − p^2 ))  Square both sides  ∴  (2p^2  − 1)^2  = [2p(√(1 − p^2 ))]^2   ∴  (2p^2  − 1)(2p^2  − 1) = 4p^2 (1 − p^2 )  ∴   4p^4  − 2p^2  − 2p^2  + 1 = 4p^2  − 4p^4   ∴   4p^4  − 2p^2  − 2p^2  + 1 + 4p^4  − 4p^2  = 0  ∴   8p^4  − 8p^2  + 1 = 0  ∴   8(p^2 )^2  − 8p^2  + 1 = 0  a = 8, b = − 8, c = 1  p^2  = ((− b ± (√(b^2  − 4ac)))/(2a))  p^2  = ((− (−8) ± (√((−8)^2  − 4(8)(1))))/(2(8)))  p^2  = ((8 ± (√(64 − 32)))/(16))  p^2  = ((8 ± (√(32)))/(16))  p^2  = 0.8536 or p^2  = 0.1465  p = ± (√(0.8536)) or p = ± (√(0.1465))  p = ± 0.9239   or p = ± 0.3828  sinx = ± 0.9239   or sinx = ± 0.3828  ∴ x = sin^(−1) (0.9239) or x = sin^(−1) (−0.9239) or x = sin^(−1) (0.3828) or x = sin^(−1) (−0.3828)  ∴ x = 67.5 or x = − 67.5 or x = 22.5 or x = − 22.5  To find other angle in the range given.  use:  x = 180n° + (−1)^n  sin^(−1) (α),      where  α = ± 67.5 or α = ± 22.5  and  n = 0, 1, 2 ...  Cancel anyone that is out of range.
$$\mathrm{sin2x}\:+\:\mathrm{cos2x}\:=\:\mathrm{0} \\ $$$$\mathrm{Note}:\:\mathrm{sin2x}\:=\:\mathrm{2sinxcosx}\:\:\mathrm{and}\:\:\mathrm{cos2x}\:=\:\mathrm{cos}^{\mathrm{2}} \mathrm{x}\:−\:\mathrm{sin}^{\mathrm{2}} \mathrm{x} \\ $$$$\mathrm{2sinxcosx}\:+\:\mathrm{cos}^{\mathrm{2}} \mathrm{x}\:−\:\mathrm{sin}^{\mathrm{2}} \mathrm{x}\:=\:\mathrm{0} \\ $$$$\mathrm{cos}^{\mathrm{2}} \mathrm{x}\:=\:\mathrm{1}\:−\:\mathrm{sin}^{\mathrm{2}} \mathrm{x} \\ $$$$\mathrm{2sinxcosx}\:+\:\mathrm{1}\:−\:\mathrm{sin}^{\mathrm{2}} \mathrm{x}\:−\:\mathrm{sin}^{\mathrm{2}} \mathrm{x}\:=\:\mathrm{0} \\ $$$$\mathrm{2sinxcosx}\:+\:\mathrm{1}\:−\:\mathrm{2sin}^{\mathrm{2}} \mathrm{x}\:=\:\mathrm{0} \\ $$$$\therefore\:\mathrm{2sin}^{\mathrm{2}} \mathrm{x}\:−\:\mathrm{2sinxcosx}\:−\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\therefore\:\mathrm{2sin}^{\mathrm{2}} \mathrm{x}\:−\:\mathrm{2sinx}\sqrt{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\:−\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\therefore\:\mathrm{2sin}^{\mathrm{2}} \mathrm{x}\:−\:\mathrm{2sinx}\sqrt{\mathrm{1}\:−\:\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\:−\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{sinx}\:=\:\mathrm{p} \\ $$$$\therefore\:\:\mathrm{2p}^{\mathrm{2}} \:−\:\mathrm{2p}\sqrt{\mathrm{1}\:−\:\mathrm{p}^{\mathrm{2}} }\:−\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\therefore\:\:\mathrm{2p}^{\mathrm{2}} \:−\:\mathrm{1}\:=\:\:\mathrm{2p}\sqrt{\mathrm{1}\:−\:\mathrm{p}^{\mathrm{2}} } \\ $$$$\mathrm{Square}\:\mathrm{both}\:\mathrm{sides} \\ $$$$\therefore\:\:\left(\mathrm{2p}^{\mathrm{2}} \:−\:\mathrm{1}\right)^{\mathrm{2}} \:=\:\left[\mathrm{2p}\sqrt{\mathrm{1}\:−\:\mathrm{p}^{\mathrm{2}} }\right]^{\mathrm{2}} \\ $$$$\therefore\:\:\left(\mathrm{2p}^{\mathrm{2}} \:−\:\mathrm{1}\right)\left(\mathrm{2p}^{\mathrm{2}} \:−\:\mathrm{1}\right)\:=\:\mathrm{4p}^{\mathrm{2}} \left(\mathrm{1}\:−\:\mathrm{p}^{\mathrm{2}} \right) \\ $$$$\therefore\:\:\:\mathrm{4p}^{\mathrm{4}} \:−\:\mathrm{2p}^{\mathrm{2}} \:−\:\mathrm{2p}^{\mathrm{2}} \:+\:\mathrm{1}\:=\:\mathrm{4p}^{\mathrm{2}} \:−\:\mathrm{4p}^{\mathrm{4}} \\ $$$$\therefore\:\:\:\mathrm{4p}^{\mathrm{4}} \:−\:\mathrm{2p}^{\mathrm{2}} \:−\:\mathrm{2p}^{\mathrm{2}} \:+\:\mathrm{1}\:+\:\mathrm{4p}^{\mathrm{4}} \:−\:\mathrm{4p}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\therefore\:\:\:\mathrm{8p}^{\mathrm{4}} \:−\:\mathrm{8p}^{\mathrm{2}} \:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\therefore\:\:\:\mathrm{8}\left(\mathrm{p}^{\mathrm{2}} \right)^{\mathrm{2}} \:−\:\mathrm{8p}^{\mathrm{2}} \:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{a}\:=\:\mathrm{8},\:\mathrm{b}\:=\:−\:\mathrm{8},\:\mathrm{c}\:=\:\mathrm{1} \\ $$$$\mathrm{p}^{\mathrm{2}} \:=\:\frac{−\:\mathrm{b}\:\pm\:\sqrt{\mathrm{b}^{\mathrm{2}} \:−\:\mathrm{4ac}}}{\mathrm{2a}} \\ $$$$\mathrm{p}^{\mathrm{2}} \:=\:\frac{−\:\left(−\mathrm{8}\right)\:\pm\:\sqrt{\left(−\mathrm{8}\right)^{\mathrm{2}} \:−\:\mathrm{4}\left(\mathrm{8}\right)\left(\mathrm{1}\right)}}{\mathrm{2}\left(\mathrm{8}\right)} \\ $$$$\mathrm{p}^{\mathrm{2}} \:=\:\frac{\mathrm{8}\:\pm\:\sqrt{\mathrm{64}\:−\:\mathrm{32}}}{\mathrm{16}} \\ $$$$\mathrm{p}^{\mathrm{2}} \:=\:\frac{\mathrm{8}\:\pm\:\sqrt{\mathrm{32}}}{\mathrm{16}} \\ $$$$\mathrm{p}^{\mathrm{2}} \:=\:\mathrm{0}.\mathrm{8536}\:\mathrm{or}\:\mathrm{p}^{\mathrm{2}} \:=\:\mathrm{0}.\mathrm{1465} \\ $$$$\mathrm{p}\:=\:\pm\:\sqrt{\mathrm{0}.\mathrm{8536}}\:\mathrm{or}\:\mathrm{p}\:=\:\pm\:\sqrt{\mathrm{0}.\mathrm{1465}} \\ $$$$\mathrm{p}\:=\:\pm\:\mathrm{0}.\mathrm{9239}\:\:\:\mathrm{or}\:\mathrm{p}\:=\:\pm\:\mathrm{0}.\mathrm{3828} \\ $$$$\mathrm{sinx}\:=\:\pm\:\mathrm{0}.\mathrm{9239}\:\:\:\mathrm{or}\:\mathrm{sinx}\:=\:\pm\:\mathrm{0}.\mathrm{3828} \\ $$$$\therefore\:\mathrm{x}\:=\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{0}.\mathrm{9239}\right)\:\mathrm{or}\:\mathrm{x}\:=\:\mathrm{sin}^{−\mathrm{1}} \left(−\mathrm{0}.\mathrm{9239}\right)\:\mathrm{or}\:\mathrm{x}\:=\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{0}.\mathrm{3828}\right)\:\mathrm{or}\:\mathrm{x}\:=\:\mathrm{sin}^{−\mathrm{1}} \left(−\mathrm{0}.\mathrm{3828}\right) \\ $$$$\therefore\:\mathrm{x}\:=\:\mathrm{67}.\mathrm{5}\:\mathrm{or}\:\mathrm{x}\:=\:−\:\mathrm{67}.\mathrm{5}\:\mathrm{or}\:\mathrm{x}\:=\:\mathrm{22}.\mathrm{5}\:\mathrm{or}\:\mathrm{x}\:=\:−\:\mathrm{22}.\mathrm{5} \\ $$$$\mathrm{To}\:\mathrm{find}\:\mathrm{other}\:\mathrm{angle}\:\mathrm{in}\:\mathrm{the}\:\mathrm{range}\:\mathrm{given}. \\ $$$$\mathrm{use}:\:\:\mathrm{x}\:=\:\mathrm{180n}°\:+\:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{sin}^{−\mathrm{1}} \left(\alpha\right),\:\:\:\:\:\:\mathrm{where}\:\:\alpha\:=\:\pm\:\mathrm{67}.\mathrm{5}\:\mathrm{or}\:\alpha\:=\:\pm\:\mathrm{22}.\mathrm{5} \\ $$$$\mathrm{and}\:\:\mathrm{n}\:=\:\mathrm{0},\:\mathrm{1},\:\mathrm{2}\:…\:\:\mathrm{Cancel}\:\mathrm{anyone}\:\mathrm{that}\:\mathrm{is}\:\mathrm{out}\:\mathrm{of}\:\mathrm{range}. \\ $$
Answered by mrW1 last updated on 16/Jun/17
(√2)((1/( (√2)))sin 2x+(1/( (√2)))cos 2x)=0  (√2)(cos (π/4) sin 2x+sin (π/4) cos 2x)=0  (√2)sin (2x+(π/4))=0  sin (2x+(π/4))=0  ⇒2x+(π/4)=kπ,k∈Z  ⇒x=(1/2)(kπ−(π/4))  within (−π,π):  x=−((5π)/8),−(π/8),((3π)/8),((7π)/8)
$$\sqrt{\mathrm{2}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{sin}\:\mathrm{2x}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{cos}\:\mathrm{2x}\right)=\mathrm{0} \\ $$$$\sqrt{\mathrm{2}}\left(\mathrm{cos}\:\frac{\pi}{\mathrm{4}}\:\mathrm{sin}\:\mathrm{2x}+\mathrm{sin}\:\frac{\pi}{\mathrm{4}}\:\mathrm{cos}\:\mathrm{2x}\right)=\mathrm{0} \\ $$$$\sqrt{\mathrm{2}}\mathrm{sin}\:\left(\mathrm{2x}+\frac{\pi}{\mathrm{4}}\right)=\mathrm{0} \\ $$$$\mathrm{sin}\:\left(\mathrm{2x}+\frac{\pi}{\mathrm{4}}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2x}+\frac{\pi}{\mathrm{4}}=\mathrm{k}\pi,\mathrm{k}\in\mathbb{Z} \\ $$$$\Rightarrow\mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{k}\pi−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\mathrm{within}\:\left(−\pi,\pi\right): \\ $$$$\mathrm{x}=−\frac{\mathrm{5}\pi}{\mathrm{8}},−\frac{\pi}{\mathrm{8}},\frac{\mathrm{3}\pi}{\mathrm{8}},\frac{\mathrm{7}\pi}{\mathrm{8}} \\ $$
Commented by myintkhaing last updated on 16/Jun/17
in 2^(nd)  line, not sin (π/2)
$${in}\:\mathrm{2}^{{nd}} \:{line},\:{not}\:{sin}\:\frac{\pi}{\mathrm{2}}\: \\ $$
Commented by mrW1 last updated on 16/Jun/17
my typo, it′s (π/4)
$$\mathrm{my}\:\mathrm{typo},\:\mathrm{it}'\mathrm{s}\:\frac{\pi}{\mathrm{4}} \\ $$
Commented by myintkhaing last updated on 16/Jun/17
sin cos +cos sin
$${sin}\:{cos}\:+{cos}\:{sin} \\ $$

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